Answer to Question #258022 in Calculus for Russell

Question

find f(x) = x2 - 2x -2 find f' (x) using the long method

Answer 1

Let $f(x) = x^2 - 2x -2$. Let us find $f' (x)$ using the long method.

$f'(x)=\lim\limits_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x} \\=\lim\limits_{\Delta x\to 0}\frac{(x+\Delta x)^2 - 2(x+\Delta x) -2-(x^2 - 2x -2)}{\Delta x} \\=\lim\limits_{\Delta x\to 0}\frac{x^2+2x\Delta x+(\Delta x)^2 - 2x-2\Delta x -2-x^2 +2x +2}{\Delta x} \\=\lim\limits_{\Delta x\to 0}\frac{2x\Delta x+(\Delta x)^2 -2\Delta x }{\Delta x} \\=\lim\limits_{\Delta x\to 0}(2x+\Delta x -2 )\\=2x-2.$